Transcript Liquids and Solids Chapter 13

Liquids and Solids
Chapter 13
Solids, Liquids, and Gases
• Solids & liquids are condensed states
– atoms, ions, molecules are close to one another
– highly incompressible
• Liquids & gases are fluids
– easily flow
• Intermolecular attractions in liquids & solids are strong
Look at Table 13-1 for a general description of characterisitics
Kinetic-Molecular Description of
Liquids and Solids
• When a gaseous sample is cooled or compressed
the rapid, random motion of the molecules
decrease. The attraction between molecules
becomes significant. When the attractive forces
overcome the reduced kinetic energies,
condensation occurs (or the gas turns to a liquid).
– In the liquid phase the particles are close. Very little
space is unoccupied (discuss).
– The particles, however, still have sufficient KE’s to
partially overcome the attractive forces. Therefore,
liquids are considered fluids and can take the shape of a
container.
Kinetic-Molecular Description of
Liquids & Solids
• Liquids that diffuse into one another are
miscible (i.e. one liquid is soluble in the other)
– Water and methanol/gasoline and motor oil
• Immiscible liquid do not diffuse into one
another (i.e. one liquid is not soluble in the
other)
– Water and hexane/water and gasoline
Kinetic-Molecular Description of
Liquids & Solids
• Cooling the liquid lowers the KE even
more. Shorter-range attractive force
become important and the liquid solidifies.
Particles in a solid cannot move freely past
one another as they can in a liquid. This is
why solids have definite shapes and
volumes. Solids are essentially
incompressible.
– If diffusion occurs in solids, it is very slow.
Intermolecular Attractive Forces
• Intermolecular attractive forces are the forces
between individual particles of a substance.
– Generally, these forces are very weak compared to
intramolecular forces (e.g. covalent and ionic bonds)
• Covalent bonding and attractive forces in H2O(l)
• Important physical properties such as boiling
points, vapor pressure, heat of vaporation, melting
points, and heat of fusion depend on the strength
of these intermolecular attractive forces.
• If intermolecular attractive forces did not exist,
solids and liquids would not exist (only gases)
Ion-Ion Interactions
• The force of attraction between two charged
particles can be determined from Coulomb’s
Law

q q
F
2
d

where q+ and q- are the charges on the particles
and d is the distance between them.
Ion-Ion Interactions
• The energy of attraction between charged particles is given
by
+
E =
q q 

Fd =
d
q q 

=
+
d2
-
d
• The energy of attraction is large for ionic compounds due
to the charged particles that are close together when
substance is a solid.
– The melting points for ionic compounds is relatively high.
• For a given substance, the separation between particles in
the solid is less than the separation in liquids.
– Energy of attraction is greater in the solid phase (generally)
Ion-Ion Interactions
• Ionic compounds that possess multiplycharged ions (e.g. Al3+ and O2-) usually have
higher melting points than ionic compounds
that posses singly-charged particles. Why
(two reasons)?
• Arrange the following ionic compounds in
the expected order of increasing melting and
boiling points.
NaF, CaO, CaF2
Dipole-Dipole Interactions
• Permanent dipole-dipole interactions occur
between polar covalent molecules because of
the attraction between - and + on different
molecules.
• Generally, these forces are not as strong as
ion-ion interactions
– Attraction between partial charges
– Dipole-dipole forces vary as 1/d4 instead of 1/d2
• Decrease faster
Dipole-Dipole Interactions
The partial positive charge on the hydrogen atoms is
attracted to the partial negative charge on the nitrogen atoms.
Note: Dipole-dipole interactions are dependent on
temperature. Why?
Hydrogen Bonding
• Hydrogen bonding is a special type of strong
dipole-dipole interaction
• Occurs between covalent molecules
containing H and of the three small, highly
electronegative elements-F, O, or N.
– One molecule most possess a H atom attached to
one of these highly electronegative atoms.
– The other molecule most possess one these highly
electronegative atoms.
Hydrogen Bonding
The partial negative charge of one molecule is attracted to
the partial positive charge of another molecule. The small
sizes of F, O, and N and their high electronegativities
concentrate the electrons around these atoms.
Hydrogen Bonding
• Typical hydrogen-bond
energies are greater than
dipole-dipole energies
– 15-20 kJ/mol for hydrogenbond energies
– 4 kJ/mol for dipole-dipole
energies
– 400 kJ/mole for ion-ion
interaction energies
• Hydrogen bonding is
responsible for the high
boiling points of water and
methanol.
Dispersion Forces
• These are attractive forces that are present
in all types of molecules.
– Dispersion forces are weak in small molecules.
– They are important at extremely small distances
which vary as 1/d7.
– They are the only attractive force present in
symmetrical nonpolar substances such as Cl2
and monatomic species.
Dispersion Forces
• Dispersion forces result from the attraction of a positively
charged nucleus to the electron cloud of another atom in
nearby molecules. As a result, temporary dipoles are
induced in the neighboring atoms or molecules.
• The magnitude of the temporary dipole increases with
increasing size of the electron cloud (or size of the
molecule). The larger electron cloud is more diffuse and
easily distorted. Adjacent molecules are polarized by the
adjacent nuclei. Polarizibility increases with increasing
sizes of molecules.
Dispersion forces are, therefore, stronger for molecules that
are larger or have more electrons.
Dispersion Forces for Argon
Dispersion forces can also exist between cations/anions and
polarizable atom or molecule. Examine Figure 13-6.
Dispersion Forces
• Some trends observed in increasing boiling
points can largely be attributed to dispersion
forces (Figure 13-5).
– CH4, SiH4, GeH4, and SnH4
– HCl, HBr, and HI
• Understand Table 13-3.
– Heat of vaporization measures the energy
required to overcome attractive forces in the
liquid.
The Liquid State
• Viscosity - the resistance to flow of a liquid.
– Generally, the higher the attractive forces in a
liquid, the greater the viscosity.
• Water versus honey or Karo syrup.
• Pentane versus dodecane
– Viscosity decreases with increasing
temperature. Why?
The Liquid State
• Surface Tension - measure of the inward
forces that must be overcome to expand the
surface area of a liquid.
– Molecules at the surface are attracted unevenly.
• Water bugs and floating razor blades (or needles)
Demo: Razor blade or needle
Liquid State
• Capillary action - ability of a liquid to rise or fall
in a tube.
– Cohesive forces – forces holding a liquid together.
– Adhesive force – forces between a liquid and another
surface
– Capillary action
• Capillary rise implies adhesive forces > cohesive forces
• Capillary fall implies cohesive forces > adhesive forces
Water is attracted to a glass capillary tube due to attractive forces
between the partial negative oxygens on the surface of the
glass and the partial positive charges on the hydrogen atoms.
Liquid State
• Capillary Action
– The meniscus of water has a concave shape due
to the strong adhesive forces between the water
molecules and the glass graduated cylinder.
– The meniscus of Hg has a convex shape
because the cohesive forces are much stronger
than the adhesive forces (Figure 13-9).
Liquid State
• Evaporation – process by which molecules escape from
the surface of a liquid.
– Only the molecules at the surface with sufficient KE will be
able to escape the attractive forces present in the liquid.
Therefore, the rate of evaporation is proportional to ________.
Liquid State
• Evaporation
– As the faster molecules leave the liquid, the
flower molecules are left behind. What will
this do to the temperature?
• This process is termed as the “cooling by
evaporation”.
DEMO: Thermometer in acetone
Liquid State
• Evaporation and condensation
– In an open container/beaker, all the water that is present
will eventually escape into the gaseous phase (Figure 1310).
– What if the beaker is sealed? What will happen? After
molecules enter the gas phase they may be recaptured by
the liquid by collisions. This process is called _______.
At some point in time, the amount of gaseous molecule
leaving the gaseous phase will equal the amount
reentering the liquid phase. This is termed as dynamic
equilibrium.
liquid





condensation
evaporation
vapor
Liquid State
• Vapor pressure – the partial pressure of vapor
molecules above the surface of a liquid at
equilibrium.
– Vapor pressure increases with temperature. Why?
Look at Figure 13-13.
– Vapor pressure decreases with increasing attractive
forces. Why? Look at Table 13-4.
• Hydrogen bonding, dipole-dipole, and dispersion forces.
DEMO: H2O and ethyl ether (add H2O to ethyl ether)
Liquid State
• As the temperature is increased, the vapor pressure
increases until the liquid boils.
• Boiling point – temperature at which the vapor
pressure equals the external pressure.
– The boiling point of H2O is less in Rexburg. Why?
• Normal boiling point – temperature at which the vapor
pressure of a liquid is equal to 1 atm.
– Boiling point at 1 atm
• What if the external pressure is lower over water?
– DEMO: Evacuate a flask filled with water
Liquid State
• Distillation – a method used to separate
components in a solution based on
differences in boiling point temperatures.
– Separation of liquids in a solution
– Purification of water (distilled water)
• Separate impurities (e.g. ions) from tap water
– Describe the basics of operation (Figure 13-14)
Of course, this can be related to vapor pressures.
The Liquid State
• Heat transfer involving liquids
• Specific heat or molar heat capacity of a liquid is the
amount of heat that must be added to 1 gram or 1 mole
to raise the temperature by 1C.
specific heat 
amount of heat in J
mass of subs tan ce  temperatur e change in
o
C
( units are
J
g C
o
How much heat is released by 200 g of H2O as it cools from
85.0oC to 40.0oC? The specific heat of water is 4.184 J/goC.
Review in Chapter 1 if needed.
)
The Liquid State
• The specific heat equation involves changing the
temperature without a change in state. For a liquid, the
temperature increases until the boiling point is acquired.
The temperature remains at the boiling point
temperature until all the liquid has been converted into
the gas.
– Boiling water is always 100C at sea level
• If the temperature is not increasing, what is all the
heat/energy being used for when the liquid is boiling?
Boiling water on a hotplate or a stove
The Liquid State
• Answer – The heat is providing the energy
necessary to break up the intermolecular attractive
interactions present in the liquid.
• Molar heat of vaporization – amount of heat that
must be added to one mole of the liquid at its
boiling point to convert it to vapor
– There is no change in temperature
How does heat of vaporization relate to intermolecular
attractive forces and vapor pressure (Table 13-5)
How does this relate to the cooling effect of perspiration?
The Liquid State
• Condensation is the reverse of
vaporization/evaporation.
• Heat of condensation is the amount of heat that
must be removed from a vapor to condense it.
liquid  heat





condensation
evaporation





condensation
evaporation
vapor
H 2 O ( l) at 100 C  2260 J
H 2 O ( g) at 100  C
– Heat is released when the substance condenses.

Liquid State
• Calculate the amount of heat necessary to convert
125 grams of water at 25.0C to steam at 100C.
• How many joules of energy must be absorbed by
500 g of H2O at 50.0oC to convert it to steam at
120oC? The molar heat of vaporization of water is
40.7 kJ/mol and the molar heat capacities of liquid
water and steam are 75.3 J/mol oC and 36.4 J/mol
oC, respectively.
• If 45.0 g of steam at 140oC is slowly bubbled into
450 g of water at 50.0oC in an insulated container,
can all the steam be condensed?
Liquid State
• Many trends in physical properties can be explained
form the strength of intermolecular attractive forces
in the liquid.
– Table 13-6 (understand these trends based on
intermolecular attractive forces)
• Arrange the following substances in order of
increasing boiling points.
– C2H6, NH3, Ar, NaCl, AsH3
– Discuss trends on Figure 13-5 (others as required)
• Generally, if a group of molecules possess the same type of
attractive forces, the boiling point increases with size/molecular
weight of the compound.
The Clausius-Clapyron Equation
• The equation relates the vapor pressure of liquids
at different temperatures.
 P2  H vap  1
1
ln   



R  T1 T2 
 P1 
– The vapor pressure is equal to 760 at the normal
boiling point.
• Appendix E and Table 13-5 produce some normal boiling
point temperatures (measured at 1 atm). These values
could be used in the equation above.
– Also, the vapor pressure at the boiling point is equal
to the atmospheric pressure
The Clausius-Clapyron Equation
• At what temperature is the vapor pressure of water
equal to the atmospheric pressure at sea level?
Would the vapor pressure of boiling water in
Rexburg be higher or lower than boiling water at
sea level? Why?
– In Rexburg, the normal atmospheric pressure is 652
torr. At what temperature does water boil in Rexburg?
– What will be the vapor pressure of ethyl alcohol at
room temperature (298 K)?
Remember to use the appropriate R.
The Solid State
• The melting point/freezing point of a substance
is the temperature at which the solid and liquid
phases coexist in equilibrium.
• The normal melting point is the melting point
at one atmosphere
– Changes in pressures have very small effects on
melting point
Heat Transfer Involving Solids
• When heat is added to a solid the temperature
increases until the melting point is acquired.
Additional heat is required to convert the solid to a
liquid (melting). Why? Heat that is added during
the melting process does not increase the
temperature. Temperature will rise after the solid is
all converted to the liquid.
Heat Transfer Involving Solids
• Molar heat of fusion (Hfus; kJ/mol) is the amount
of heat required to melt one mole of a solid at its
melting point. This can be converted to heat of
fusion which is (kJ/g)
– Heat of fusion depends on _________(Table 13-7)
• Molar heat of solidification/crystallization is equal
in magnitude (but opposite sign) to the molar heat
of fusion.

o
H 2 O ( s) at 0 o C  6 .02 kJ / mol 
 H 2 O ( l) at 0 C
melting
freezing
– 6.02 kJ is absorbed when 1 mole is melted and 6.02 kJ
is released when 1 mole is frozen/solidifies.
Heat Transfer Involving Solids
Heat Transfer Involving Solids
• Calculate the amount of heat required to
convert 150.0 g of ice at -10.0oC to water at
40.0oC. The specific heat of ice is 2.09 J/goC
• Calculate the amount of heat required to
convert 75.0 g of solid ethanol at –117.0C to
gaseous ethanol at 95.0C.
– Table 13-5, 13-7 and and Appendix E
Sublimation and Vapor Pressures
of Solids
• Sublimation is the process by which a solid forms
a gas (vaporizes) without passing through the
liquid phase.
– Dry ice (CO2)
• Deposition is the reverse; process by which a
vapor forms a solid without passing through the
liquid phase.
– Discussion with chemical vapor deposition
Note: Solids have vapor pressures, but they are
generally very small
Phase Diagrams (P versus T)
• Phase diagrams illustrate a particular phase
or state that is present under specific
temperature-pressure conditions.
– Temperature is on the y-axis and pressure is on
the x-axis.
• Using a phase diagram, changes in state can
be determined when changing pressure
and/or temperature.
Phase Diagrams (P versus T)
• Phase boundaries
– Line AC – liquid and gas
phases coexist at equilibrium
– Line AB – liquid and solid
phases coexist at equilibrium
– Line AD – gas and solid
phases coexist at equilibrium
– Triple point (A) – at this T and
P all three phases coexist at
equilibrium
Phase Diagrams (P versus T)
• Notice the negative AB
line slope. This is unique
to H2O. What happens
upon compressing a solid
(up from point J)
• Is there a pressure where
sublimation could occur?
• Travel a few other lines.
Phase Diagrams (P versus T)
• When the temperature is
increased at normal pressures
the solid goes directly to the
gas phase. At what pressure
would the solid go through
the liquid phase.
• Notice the positive slope on
the solid-liquid phase
boundary. What does this
mean?
Amorphous and Crystalline Solids
• Amorphous solids do not possess well-ordered
structures.
– Some can also be characterized as glasses since they
flow slowly.
• Windows
– Melting extend over a large range.
– Rubber, plastics, and amorphous sulfur
• Crystalline solids have well-defined structures
consisting of repeating units.
– Repeating unit can be observed upon shattering the
crystal.
– Possess distinct melting points.
Structures of Crystals
• Crystals contain regularly repeating structures.
• Unit cell is the smallest repeating unit of a crystal. A
unit cell is the fundamental box that describes the
arrangement of particles in a crystal. These unit cells
are stacked in three dimensions to produce a crystal.
The arrangement of these unit cells fit into one of
seven crystal systems. (Table 13-9).
– Crystals have the same symmetry as the unit cells since the
crystals are built from multiple units of these cells.
Structures of Crystals
Look on page 513. Notice that the structures of the unit
cells are very similar to the actual structure of the
crystal. Why?
Structures of Crystals
• Creating the crystal from the unit cells
– Consider the corner of a unit cell as a lattice
point. In three dimensions, this lattice point is
shared by eight unit cells. If an object were
present at this lattice point, it would be equally
shared by all eight unit cells (1/8 in each). How
many objects would be in each unit cell?
DEMO: Use unit cell building blocks to
illustrate.
Structures of Crystals
• In a simple or primitive lattice structure (discussed
previously), only the corners were occupied by
objects. In other crystal types, objects may also
occupy other positions in the unit cell.
– Simple cubic (no additional objects)
– Body-centered cubic (bcc) – another object occupies
the center of the unit cell
– Face-centered cubic (fcc) – another six object occupies
the middle of each of the six square faces of the cube
Structures of Crystals
How many objects/atoms per unit cell for each crystal
type? Discuss
Look at Figure 13-24.
Bonding in Solids
• Categories of crystalline solids
– Metallic solids
– Ionic solids
– Molecular solids
– Covalent solids
Table 13-10 summarizes properties of each
category type.
Metallic Solids
• Positively charged nuclei are surrounded by a sea
of delocalized valence electrons. This is the
reason why most metals are good conductors.
– The nuclei occupy lattice sites
• Lattice types for metals
– Body-centered cubic (bcc)
– Face-centered cubic (fcc)
– Hexagonal close-packed (hcp)
Examples: Li, K, Ca, and Au
Metallic Solids
• Obtaining the close-packed structures (fcc and
hcp)
– Hexagonal close packed structure has an ABA
arrangement.
• The different letters correspond to different planes (Figure 1327a).
– Face-centered cubic (or cubic close-packed) structure
has an ABC arrangement.
• Figure 13-27b
For the the close-packed structures approximately 74% of the
volume is occupied.
The body-centered cubic structures has much less of its volume
occupied by metal spheres.
Metallic Solids
• Manganese has a simple cubic unit cell. The atomic radius
is 3.15 Å. What is the shortest distance between
neighboring Mn atoms? How many nearest neighbors
does each atoms have?
• Nickel crystals are face-centered cubic. The radius of the
nickel atom in the metal is 1.24Å. What is the distance
between centers of the two closest Ni atoms. What is the
length of the cell edge. How many nearest neighbors does
each atoms have?
• Calculate the density of metallic nickel. Determine the
percentage of space that is occupied by the nickel atoms.
Metallic Solids
• A group IVA element with a density of
11.35 g/cm3 crystallizes in a face-centered
cubic lattice whose unit cell edge length is
4.95 A. Calculate the element’s atomic
weight. What is the atomic radius of this
element?
Ionic Solids
• Most salts crystallize as ionic solids with ions
occupying the unit cell. The most common
example is sodium chloride, which has a
face-centered cubic arrangement.
– Many other salts that have the same charge on
both the anion and cation have the same fcc
arrangement (LiCl and MgO).
• Even though the solid compound possesses
charge, it does not conduct electricity. Why?
Ionic Solids
• How many Cl- and Na+ ions are in the fcc unit
cell? Remember, a sodium ion is in the center.
– Illustrate models and look at 13-13 on demo CD.
Sodium iodide crystallizes in the fcc structure (like NaCl).
The I- ion radius is 2.20 Å. The I- ions at the corners of
the unit cell are in contact with those at the centers of
the faces. Determine the length of the unit cell.
Calculate the radius of the Na+ ion assuming anioncation contact.
Molecular Solids
• Molecules occupy the lattice positions of the unit
cell.
– There are covalent bond within the molecules but only
intermolecular attractive forces between molecules.
What are the types?
• These solids tend to have lower melting points
since the forces holding the molecules together are
weaker.
• Look at page 527 in book and 13-14 on demo CD.
Covalent Solids (‘network solids’)
• These can be considered giant molecules
with the atom bond covalently in a
crystalline network.
– Structures are usually very hard with high
melting points. Most are poor thermal and
electrical conductors.
– Examples: Diamond, graphite, and quartz
• What is the bonding in diamond (hydridization) and
graphite?
13-15 on demo. CD
Band Theory of Metals
• Recall that the bonding in metals is due to
delocalized, mobile electrons that belong to the
solid as a whole.
– Responsible for the conduction (electrical and thermal)
of most metals.
• Band theory of metals is used to explain properties
of metals and other materials.
– According to the MO theory, atomic orbitals overlap to
produce a set of molecular orbitals. The number of
generated molecular orbitals is equal to the number of
overlapping atomic orbitals. There is a very large
number of atomic orbitals in a metal!!!
Band Theory of Metals
• In a Na metal, the 3s atomic orbitals overlap to
produce a very large set of molecular orbitals that
are very closely spaced in energy. These closely
spaced orbitals are called a band of orbitals. Since
there is only one electron in the 3s atomic orbital,
the molecular orbitals are only half filled. Only a
small amount of energy is needed for the highest
energy electrons in the 3s band to jump into a
vacant orbital at a slightly higher energy.
– As a result, electrons will flow with an applied field.
Band Theory of Metals
What if the 3s band is filled? In the Group IIA metals,
the 3s band is filled, but the metals still conduct.
Band Theory of Metals
• For Mg, the 3s and 3p
band overlap since there
is a range of energies for
the molecular orbitals.
– Without this overlap Mg
metal would not conduct.
• The highest energy
electrons are able to
move into the vacant
orbitals in the 3p band.
Band Theory of Metals
• Why don’t some materials conduct
electricity?
– The highest energy electrons in insulators occupy
filled bands of molecular orbitals that are well
separated from the lowest empty band.
– For semiconductors, the filled bands are only
slightly below the empty bands.
• Generally, increasing the temperature or ‘doping’ the
material will produce conduction (page 518-519).
Band Theory of Metals